BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 59
Joined: Thu May 05, 2016 4:20 am
Thanked: 3 times
by Joy Shaha » Fri Jul 15, 2016 5:26 am
Q. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Fri Jul 15, 2016 6:33 am
Hi Joy Shaha,

This question can be solved in a couple of different ways. I'm going to approach this with permutation/probability math.

We're asked for the probability of pulling a red and a white (in either order) from a group of 3 red, 4 black and 2 white balls (with replacement). There are two ways to accomplish this:

(White first)(Red second) = (2/9)(3/9) = 6/81

(Red first)(White second) = (3/9)(2/9) = 6/81

Total = 6/81 + 6/81 = 12/81 = 4/27

Final Answer: D

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 410
Joined: Fri Mar 13, 2015 3:36 amFollowed by: 8 members
Location: Worldwide
Thanked: 120 times
GMAT Score:770

by OptimusPrep » Sat Jul 16, 2016 11:43 am
Joy Shaha wrote:Q. A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
3R, 4B, 2W balls.

P(1R, 1W) = (3/9)*(2/9) = 1/3*2/9 = 2/27
Now the Red and the While balls can be drawn in any order
Hence probability = 2*2/27 = 4/27

Correct Option: D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Wed Jul 20, 2016 10:01 pm
You could also do

1 - P(Both Red) - P(Both White) - P(Either Black)

or

1 - (3/9)*(2/9) - (2/9)*(1/9) - (1 - (5/9)*(4/9))

or

1 - 6/81 - 2/81 - 61/81

or

12/81 => 4/27
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1